Question

Find the equation of the circle which has its centre on the line $y=2$ and which passes through the points $(2,0)$ and $(4,0)$

Answer 1

The circle has centre on line $y=2$

Let the centre be $(x_1,y_1)$

$\Rightarrow y_1=2$

circle passses through $(2,0)$ and $(4,0)$

We know $OA=OB=R$

$PA=PB$(perpendicular from centre bisects the chord)

But $PA+PB=\sqrt{(4-2{)}^2+D^2}=2$

$2PA=2$

$\Rightarrow PA=1$

$\therefore$ Coordinates of $P=(2+1,0)\Rightarrow (3,0)$

Since OP is parallel to y-axis.

centre lies along the line $x=3$

we also know it lies along the line $y=2$

$\therefore$ centre $=(3,2)$

$\Rightarrow$ Radius $=\sqrt{(3-2{)}^2+(2-0{)}^2}$

$=\sqrt{5}$

Equation of circle is $(x-3{)}^2+(y-2{)}^2=5$.